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anonymous

  • one year ago

How to exchange the integral order, i can't solve this problem. ∫∫ xy dxdy, D = {(x,y) | -2 <= y <= 4, (y^2-6)/2 <= x <= y+1}

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  1. phi
    • one year ago
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    Here is a plot of the region, showing a "typical" rectangle as we integrate x first.

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  2. phi
    • one year ago
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    If we integrate dy first, then we evidently have to break the problem into two regions. from x=-3 to -1, and then from x=-1 to 5

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  3. phi
    • one year ago
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    from the lower limit (in the original problem) \[ x= y^2 -6 \] we get \[ y= \pm \sqrt{x+6} \] in the region between x=-3 and x=-1, evidently the lower limit is \( y=- \sqrt{x+6}\) and the upper limit is \( y= \sqrt{x+6} \) and we have \[ \int_{-3}^{-1} \int_{- \sqrt{x+6}}^{+ \sqrt{x+6}} x y\ dy \ dx \] once we get to x=-1 we change the lower limit on y. we originally had x=y+1, from which we get y= x-1 the second iterated integral is \[ \int_{-1}^{5} \int_{x-1}^{+ \sqrt{x+6}} x y\ dy \ dx \]

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